Question

What is the $pH$ of the solution when $0.20$ mole of $HCl$ is added to one litre of a solution containing
(i) $1 M$ each of acetic acid and acetate ion,
(ii) $0.1 M$ each of acetic acid and acetate ion?
Assume the total volume is one litre.
$K_{a}$ for acetic acid $=1.8 \times 10^{-5}$

Answer 1

(i) $0.20$ mole $HCl$ will neutralise $0.20$ mole $CH _{3} COONa$, producing $0.20 \,mol$ $CH _{3} COOH$.
Therefore, in the solution moles of $CH _{3} COOH =1.20$
Moles of $CH_3COONa = 0.80$
$pH = p K_{a}+\log \frac{[\text { Salt }]}{[\text { Acid }]}$
$=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{(0.80)}{(1.20)}=4.56$

Now, the solution has $0.2$ mole acetic acid and $0.1$ mole $HCl$. Due to presence of $HCl$, ionisation of $CH _{3} COOH$ can be ignored (common ion effect) and $H$ in solution is mainly due to $HCl$.
${\left[H^{+}\right]=0.10 }$
$pH =-\log (0.10)=1.0$